从分配给变量PHP的下拉列表中获取选择 [英] get a selection from a dropdown list assigned to a variable PHP

问题描述

我知道我很简单，但我只是可以找到告诉我如何做的资源。

I know I this is simple, but i just can;t find the resource that tells me how to do it.

所以我的代码如下： / p>

SO my code is as follows:

session_start();
$wquery=
"select week_num,week_name
from stats_week
where season=$this_season
order by week_num";
    $wresult=mysql_query($wquery);


print'<form action="changeplayer_2.php" method="post">';


        print"<select name='Week_select'> <br>";
        while ($wrow=mysql_fetch_array($wresult))
    { 
    print '<option value="'.$wrow['week_num'].'">'.'week '.$wrow['week_num'].'     '.$wrow['week_name'].'</option><br>\n';  
    }

print "</select><br><br>";#

    print'<button type="submit" >Next</button>';


    print"</form>";

所以我做一个选择：
我想要的选择最终在：$ _SESSION ['week']

So I am making a selection: I want that selection to end up in: $_SESSION['week']

推荐答案

内部文件changeplayer_2.php中，您需要将POSTed值加载到会话：

Inside file "changeplayer_2.php" you'll want to load the POSTed value to the session:

session_start();
$_SESSION['week'] = $_POST['Week_select'];

每当你想在关键'周'获得会话的值时，只需执行以下操作：

Whenever you want to get the value of the session at key 'week', simply do:

$ weekValue = isset（$ _ SESSION ['week']）？ $ _SESSION ['week']：false;

但是，如果您在任何脚本中使用会话变量，请确保您开始会话使用 session_start（）。

But if you do use the session variable on any script, make sure you start the session using session_start().

要销毁会话变量，您可以执行以下操作：

To destroy the session variable, you can do:

session_start();
unset($_SESSION['week']);

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